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📅 May 4, 2026 👤 scraface

May 04, 2026

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14\mathbf{v} = \langle 8, 2, -4 \rangle \quad \Rightarrow \quad \mathb...
Thus, the vector is $ \boxed{\left\langle \dfrac{4}{7}, \dfrac{1}{7},...
Question: Among all roots of the polynomial $ z^6 - z^4 + z^2 - 1 = 0...
Solution: Consider the polynomial:
z^6 - z^4 + z^2 - 1 = 0
So the roots satisfy either $ z^2 - 1 = 0 $ or $ z^4 + 1 = 0 $.
Roots from $ z^2 - 1 = 0 $:
$ z = \pm 1 $, both real, so imaginary part is 0.
Roots from $ z^4 = -1 $:
Write $ -1 = e^{i\pi} $, so:

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